I have a question about this theory...
If I have a 350 Chevy at 6500 and 14lbs of boost with a 72% eff. Compressor and
a VE 100% of I get 75Lbs per minute of air non intercooled.
If I now use the same engine and add a air to water intercooler with a 45 dgree
water and an Eff or 90% which is common I get 99 Lbs of air per minute and a
corresponding increase in power. I would also probably need a larger turbo as
well to provide the 32% increase in lbs of air that would be too large for the
non-intercooled version.
If instead of using the intercooler I used the water injection and the turbo
sized for the non intercooled version where do the pounds of air come from??
Don't I have 75 lbs per minute of colder air?? How would anyone propose to keep
the larger turbo out of surge if the water injection had a bad day or were wrong
on the btu calculation? The change in charge air temp to go from non intercooled
with 223 degree air to intercooled with 54 degree air is rather substanial How
much water would you need??
The same engine at 30 lbs of boost is 75 dgree air intercooled and 342 non
intercooled... The chemical part I personally have no clue about just looking at
mass airflow to begin with. I do have Heywoodss writings so will take a further
look at that.
Dave Dahlgren
"Waldron, James" wrote:
>
> This would seem to argue strongly for water injection into the intake
> air stream over other cooling methods for supercharged engines.
> (Did I get it?) I assume the 36% water injection is by mass?
> Using an ice water/air intercooler could supply a water source. Best of
> both worlds? You are correct that you have to haul the extra mass of
> the water around.
>
> Thanks! Very interesting.
>
> Jim W.
>
> -----Original Message-----
> From: DrMayf
> I ran across a short write up on combustion chamber chemistry ...
> http://www.teknett.com/pwp/drmayf/injchem.htm
>
> mayf, the red necked ignorant desert rat in PAH rump
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